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In a GP of positive terms, any term is equal to one-third of the sum of next two terms. What is the common ratio of the GP?
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Verified Answer
The correct answer is:
$\frac{\sqrt{13}-1}{2}$
Let a, ar and ar $^{2}$ be three positive terms of G.P.
According to question,
$\mathrm{a}=\frac{1}{3}\left(\mathrm{ar}+\mathrm{ar}^{2}\right)$
$\Rightarrow 3=\mathrm{r}+\mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}+\mathrm{r}-3=0$
$\Rightarrow \mathrm{r}=\frac{-1 \pm \sqrt{1+4 \times 3}}{2}$
$\Rightarrow \mathrm{r}=\frac{-1 \pm \sqrt{13}}{2}=\frac{\sqrt{13}-1}{2},-\left(\frac{1+\sqrt{13}}{2}\right)$
Since, $r$ can not be negative.
$\therefore \quad r=\frac{\sqrt{13}-1}{2}$
According to question,
$\mathrm{a}=\frac{1}{3}\left(\mathrm{ar}+\mathrm{ar}^{2}\right)$
$\Rightarrow 3=\mathrm{r}+\mathrm{r}^{2}$
$\Rightarrow \mathrm{r}^{2}+\mathrm{r}-3=0$
$\Rightarrow \mathrm{r}=\frac{-1 \pm \sqrt{1+4 \times 3}}{2}$
$\Rightarrow \mathrm{r}=\frac{-1 \pm \sqrt{13}}{2}=\frac{\sqrt{13}-1}{2},-\left(\frac{1+\sqrt{13}}{2}\right)$
Since, $r$ can not be negative.
$\therefore \quad r=\frac{\sqrt{13}-1}{2}$
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